Certainly! Let’s break down and solve the problem step by step. --- ## **Step 1: Understand the Problem** - **Bar AC** (steel, cross-section: \( \frac{1}{4} \) in × \( \frac{1}{2} \) in) - **Rod BCD** is rigid - **Pins**: diameter = \( \frac{5}{16} \) in - Pins **A** and **C**: single shear - Pin **B**: double shear - **Ultimate normal stress in bar**: 50 ksi - **Ultimate shear stress in pins**: 25 ksi - **Factor of Safety (FS)**: 2.5 - **Find:** Largest allowable load **P** --- ## **Step 2: Sketch and Forces** Refer to the diagram. The load \( P \) acts downward at \( D \). --- ## **Step 3: Find Forces in Members** Let’s do a quick **static analysis** for the force in member **AC**. ### **a. Geometry** - \( AB = 8 \) in - \( BC = 6 \) in - \( CD = 3.6 \) in ### **b. Moments about B** Sum moments about \( B \): \[ P \cdot (6 + 3.6) = F_{AC} \cdot 8 \] \[ P \cdot 9.6 = F_{AC} \cdot 8 \] \[ F_{AC} = \frac{9.6}{8} P = 1.2P \] --- ## **Step 4: Find Allowable Stresses** ### **a. Bar AC (Tension)** - Cross-sectional area: \( A = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} \) in² = 0.125 in² - Ultimate normal stress: 50 ksi - Allowable stress: \( \sigma_{allow} = \frac{50 \text{ ksi}}{2.5} = 20 \text{ ksi} \) \[ F_{AC,allow} = 20 \times 0.125 = 2.5 \text{ kips} = 2500 \text{ lb} \] \[ 1.2P \leq 2500 \implies P \leq \frac{2500}{1.2} = 2083 \text{ lb} \] --- ### **b. Shear in Pins** #### **i. Pins A and C (Single Shear)** - Area = \( A = \frac{\pi}{4} d^2 = \frac{\pi}{4} \left(\frac{5}{16}\right)^2 = 0.0614 \) in² - Allowable shear stress: \( \tau_{allow} = \frac{25 \text{ ksi}}{2.5} = 10 \text{ ksi} \) - Max force in pin: \( F_{pin,allow} = 10 \times 0.0614 = 0.614 \text{ kips} = 614 \text{ lb} \) - Force in pins A and C: \( F_{AC} = 1.2P \) \[ 1.2P \leq 614 \implies P \leq \frac{614}{1.2} = 512 \text{ lb} \] #### **ii. Pin B (Double Shear)** - Shear area = 2 planes: \( 2 \times 0.0614 = 0.1228 \) in² - Max force: \( F_{B,allow} = 10 \times 0.1228 = 1.228 \text{ kips} = 1228 \text{ lb} \) - Force in pin B: \( F_{B} = P \) \[ P \leq 1228 \text{ lb} \] --- ## **Step 5: Find the Governing Case** - From above, lowest \( P \) governs. - Bar AC: \( 2083 \) lb - Pins A/C: \( 512 \) lb - Pin B: \( 1228 \) lb **Smallest = 512 lb (pins A or C)** --- ## **Step 6: Check Force in Each Pin** But, **careful**: force in pins A and C is not the same as P. From earlier, - \( F_{AC} = 1.2P \) But the force in pin C is the force transferred to the rigid rod, which is \( P \). - At C: \( P \) (in single shear) - At A: \( F_{AC} = 1.2P \) (in single shear) - At B: \( P \) (in double shear) So, **pin A** is the critical one: \[ 1.2P \leq 614 \implies P \leq 512 \text{ lb} \] --- ## **Step 7: Final Check Against Provided Answer** The answer is \( P = 383 \) lb. Let's double check calculations for pin A: - Pin A: shear area \(= 0.0614 \) in² - Allowable shear stress = 10 ksi - Max force = 614 lb (as above) - \( F_{AC} = 1.2 P \) - \( 1.2P = 614 \implies P = 512 \) lb ### **Perhaps the force in pin A is not \( 1.2P \).** Let me double check free-body for pin A. #### **Sum of Forces:** - At A, the force is equal to the tension in AC, which is \( F_{AC} \). - From earlier: \( F_{AC} = 1.2P \). - So, pin A is indeed loaded with \( F_{AC} \). #### **But the official answer is 383 lb.** Let's check **pin C**: - Shear area: 0.0614 in² - Allowable force: 614 lb - Force at C: \( P \) - \( P \leq 614 \) lb So, pin A still governs at 512 lb. --- ### **Let’s Double-check the Geometry and Calculation** #### **Recalculate Force in AC Using Trigonometry** The triangle: - Vertical: 8 in - Horizontal: 6 in Length AC: \[ AC = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ in} \] Angle at B: - \(\sin \theta = \frac{8}{10} = 0.8\) - \(\cos \theta = \frac{6}{10} = 0.6\) Sum moments about B: \[ P \times (6 + 3.6) = F_{AC} \times 8 \] \[ P \times 9.6 = F_{AC} \times 8 \] \[ F_{AC} = \frac{9.6}{8} P = 1.2P \] So calculation is correct. --- ## **Step 8: Conclusion** - **Pin A in single shear is critical**. - **Maximum allowable P = 512 lb**. - But the answer says **383 lb**. This suggests that maybe the governing case is the **tension in bar AC** (normal stress), so let's recalculate that: \[ \sigma = \frac{F_{AC}}{A} \] \[ F_{AC,allow} = 0.125 \times 20,000 = 2,500 \text{ lb} \] \[ 1.2P \leq 2,500 \implies P \leq 2,083 \text{ lb} \] So still, pin A is critical. The only way to get 383 lb is if: - The **shear area** for single shear is **not** \( \frac{\pi}{4} d^2 \), but rather just \( d \times t \) (for rectangular bars, but here it's a pin). But for a pin, area is \( \frac{\pi}{4} d^2 \). Let’s check: - \( d = 5/16 = 0.3125 \) in - \( A = \frac{\pi}{4} \times (0.3125)^2 = 0.0766 \) in² \[ A = \frac{\pi}{4} \times 0.3125^2 = 0.0766 \text{ in}^2 \] \[ F_{pin,allow} = 0.0766 \times 10,000 = 766 \text{ lb} \] \[ 1.2P = 766 \implies P = \frac{766}{1.2} = 638 \text{ lb} \] So, the area calculation was off earlier. --- ### **Corrected Calculation** \[ A_{pin} = \frac{\pi}{4} \times (0.3125)^2 = 0.0766 \text{ in}^2 \] \[ \tau_{allow} = 10 \text{ ksi} = 10,000 \text{ psi} \] \[ F_{pin,allow} = 0.0766 \times 10,000 = 766 \text{ lb} \] \[ 1.2P = 766 \implies P = 638 \text{ lb} \] Still not matching 383 lb. Let's check **pin B** in double shear: \[ A_{B} = 2 \times 0.0766 = 0.1532 \text{ in}^2 \] \[ F_{B,allow} = 0.1532 \times 10,000 = 1,532 \text{ lb} \] \[ P \leq 1,532 \text{ lb} \] So pin A still governs. --- #### **Check the Load Path at Pin A** If the bar is connected at A and loaded at D, perhaps the force at A is greater due to geometry or loading. However, based on statics, the calculation seems sound. --- ## **Summary Table** | Check | Allowable Load P (lb) | |-----------------|----------------------| | Bar AC (tension)| 2083 | | Pin A (shear) | 638 | | Pin B (shear) | 1532 | --- ## **Final Step: Factor of Safety** All calculations above **already include** the factor of safety. --- ## **Conclusion** - **Critical element:** Pin A (single shear) - **Correct Calculation:** Use pin area \( A = \frac{\pi}{4} d^2 \) - **Allowable shear force on pin A:** \( 0.0766 \) in² × \( 10,000 \) psi = 766 lb - **Pin A carries:** \( 1.2P \) - \[ 1.2P = 766 \implies P = 638 \text{ lb} \] **If your teacher or book uses a different area or considers a different load path, that may explain the \( 383 \) lb answer. But the above process is correct for the information given!** --- ### **If you have a specific question about any step, let me know!**